Question

The quadratic equations $x^{2}-6 x+a=0$ and $x^{2}-c x$ $+6=0$ have one root in common. The other roots of the first and second equations are integers in the ratio $4: 3$. Then the common root is

• A. 1
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (D)

Let $\alpha$ and $4 \beta$ be roots of $x^{2}-6 x+a=0$ and $\alpha, 3 \beta$ be the roots of $x^{2}-c x+6=0$, then
$\alpha+4 \beta=6$ and $4 \alpha \beta=a$
$\alpha+3 \beta=c$ and $3 \alpha \beta=6$
We get $\alpha \beta=2 $
$\Rightarrow a=8$
So the first equation is $x^{2}-6 x+8=0$
$ \Rightarrow x=2,4$
If $\alpha=2$ and $4 \beta=4$ then $3 \beta=3$
If $\alpha=4$ and $4 \beta=2$, then $3 \beta=3 / 2$
(non-integer)
$\therefore $ common root is $x=2$.